使用rfind重载，接受搜索位置pos参数，并为其传递0:

std::string s = "tititoto";
if (s.rfind("titi", 0) == 0) { // pos=0 limits the search to the prefix
  // s starts with prefix
}

谁还需要别的东西?纯STL !

许多人将其误解为“在整个字符串中反向搜索前缀”。这将给出错误的结果(例如string("tititito").rfind("titi")返回2，因此当与== 0进行比较时将返回false)，并且它将是低效的(查看整个字符串而不是只查看开头)。但是它没有这样做，因为它将pos参数传递为0，这将搜索限制在该位置或更早的位置进行匹配。例如:

std::string test = "0123123";
size_t match1 = test.rfind("123");    // returns 4 (rightmost match)
size_t match2 = test.rfind("123", 2); // returns 1 (skipped over later match)
size_t match3 = test.rfind("123", 0); // returns std::string::npos (i.e. not found)

我使用std::string::compare包装在实用方法如下:

static bool startsWith(const string& s, const string& prefix) {
    return s.size() >= prefix.size() && s.compare(0, prefix.size(), prefix) == 0;
}

if(boost::starts_with(string_to_search, string_to_look_for))
    intval = boost::lexical_cast<int>(string_to_search.substr(string_to_look_for.length()));

这是完全未经测试的。原理与Python相同。需要提高。StringAlgo和boost。lexicalcast。

检查字符串是否以另一个字符串开头，然后获取第一个字符串的子字符串('slice')并使用词法转换。

c++ 20更新:

使用std:: string:: starts_with

https://en.cppreference.com/w/cpp/string/basic_string/starts_with

std::string str_value = /* smthg */;
const auto starts_with_foo = str_value.starts_with(std::string_view{"foo"});

还没有人使用STL算法/错配函数。如果返回true, prefix是'toCheck'的前缀:

std::mismatch(prefix.begin(), prefix.end(), toCheck.begin()).first == prefix.end()

完整的例子:

#include <algorithm>
#include <string>
#include <iostream>

int main(int argc, char** argv) {
    if (argc != 3) {
        std::cerr << "Usage: " << argv[0] << " prefix string" << std::endl
                  << "Will print true if 'prefix' is a prefix of string" << std::endl;
        return -1;
    }
    std::string prefix(argv[1]);
    std::string toCheck(argv[2]);
    if (prefix.length() > toCheck.length()) {
        std::cerr << "Usage: " << argv[0] << " prefix string" << std::endl
                  << "'prefix' is longer than 'string'" <<  std::endl;
        return 2;
    }
    if (std::mismatch(prefix.begin(), prefix.end(), toCheck.begin()).first == prefix.end()) {
        std::cout << '"' << prefix << '"' << " is a prefix of " << '"' << toCheck << '"' << std::endl;
        return 0;
    } else {
        std::cout << '"' << prefix << '"' << " is NOT a prefix of " << '"' << toCheck << '"' << std::endl;
        return 1;
    }
}

编辑:

正如@James T. Huggett所建议的，std::equal更适合回答这个问题:a是B的前缀吗?并且是略短的代码:

std::equal(prefix.begin(), prefix.end(), toCheck.begin())

完整的例子:

#include <algorithm>
#include <string>
#include <iostream>

int main(int argc, char **argv) {
  if (argc != 3) {
    std::cerr << "Usage: " << argv[0] << " prefix string" << std::endl
              << "Will print true if 'prefix' is a prefix of string"
              << std::endl;
    return -1;
  }
  std::string prefix(argv[1]);
  std::string toCheck(argv[2]);
  if (prefix.length() > toCheck.length()) {
    std::cerr << "Usage: " << argv[0] << " prefix string" << std::endl
              << "'prefix' is longer than 'string'" << std::endl;
    return 2;
  }
  if (std::equal(prefix.begin(), prefix.end(), toCheck.begin())) {
    std::cout << '"' << prefix << '"' << " is a prefix of " << '"' << toCheck
              << '"' << std::endl;
    return 0;
  } else {
    std::cout << '"' << prefix << '"' << " is NOT a prefix of " << '"'
              << toCheck << '"' << std::endl;
    return 1;
  }
}

在c++ 20中，starts_with作为std::string的成员函数可用，定义为:

constexpr bool starts_with(string_view sv) const noexcept;

constexpr bool starts_with(CharT c) const noexcept;

constexpr bool starts_with(const CharT* s) const;

所以你的代码可以是这样的:

std::string s{argv[1]};

if (s.starts_with("--foo="))